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1: Single-Factor ANOVA and the F Test

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15.1

Mean of

Sample 1

Mean of

Sample 2

Single-Factor ANOVA and the F Test

705

Mean of

Sample 3

(a)

FIGURE 15.1

Two possible ANOVA data sets when

three populations are under investigation: green circle 5 observation from

Population 1; orange circle 5 observation from Population 2; blue circle 5

observation from Population 3.

Mean of

Sample 1

Mean of

Sample 2

Mean of

Sample 3

(b)

After looking at the data in Figure 15.1(a), almost anyone would readily agree

that the claim m1 ϭ m2 ϭ m3 appears to be false. Not only are the three sample means

different, but also the three samples are clearly separated. In other words, differences

between the three sample means are quite large relative to the variability within each

sample. (If all data sets gave such obvious messages, statisticians would not be in such

great demand!)

The situation pictured in Figure 15.1(b) is much less clear-cut. The sample

means are as different as they were in the first data set, but now there is considerable

overlap among the three samples. The separation between sample means here can

plausibly be attributed to substantial variability in the populations (and therefore the

samples) rather than to differences between m1, m2, and m3. The phrase analysis of

variance comes from the idea of analyzing variability in the data to see how much can

be attributed to differences in the m’s and how much is due to variability in the individual populations. In Figure 15.1(a), the within-sample variability is small relative

to the between-sample variability, whereas in Figure 15.1(b), a great deal more of the

total variability is due to variation within each sample. If differences between the

sample means can be explained by within-sample variability, there is no compelling

reason to reject H0.

Notations and Assumptions

Notation in single-factor ANOVA is a natural extension of the notation used in

Chapter 11 for comparing two population or treatment means.

ANOVA Notation

k 5 number of populations or treatments being compared

Population or treatment

Population or treatment mean

Population or treatment variance

Sample size

Sample mean

Sample variance

1

m1

s21

n1

x1

s 21

2

m2

s22

n2

x2

s 22

p

p

p

p

p

p

k

mk

s2k

nk

xk

s 2k

N 5 n1 1 n2 1 % 1 nk (the total number of observations in the data set)

T 5 grand total 5 sum of all N observations in the data set 5 n1 x1 1 n2 x2 1 % 1 nk xk

T

x 5 grand mean 5

N

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706

Chapter 15 Analysis of Variance

A decision between H0 and Ha is based on examining the x values to see whether

observed discrepancies are small enough to be attributable simply to sampling variability or whether an alternative explanation for the differences is more plausible.

David Scharf/Getty Images

EXAMPLE 15.1

Activated platelet

TABLE 1 5. 1

Group

Number

1

2

3

4

An Indicator of Heart Attack Risk

The article “Could Mean Platelet Volume Be a Predictive Marker for Acute Myocardial

Infarction?” (Medical Science Monitor [2005]: 387-392) described an experiment in

which four groups of patients seeking treatment for chest pain were compared with respect to mean platelet volume (MPV, measured in fL). The four groups considered were

based on the clinical diagnosis and were (1) noncardiac chest pain, (2) stable angina

pectoris, (3) unstable angina pectoris, and (4) myocardial infarction (heart attack). The

purpose of the study was to determine if the mean MPV differed for the four groups,

and in particular if the mean MPV was different for the heart attack group, because then

MPV could be used as an indicator of heart attack risk and an antiplatelet treatment

could be administered in a timely fashion, potentially reducing the risk of heart attack.

To carry out this study, patients seen for chest pain were divided into groups

according to diagnosis. The researchers then selected a random sample of 35 from

each of the resulting k ϭ 4 groups. The researchers believed that this sampling process

would result in samples that were representative of the four populations of interest

and that could be regarded as if they were random samples from these four populations. Table 15.1 presents summary values given in the paper.

Summary Values for MPV Data of Example 15.1

Sample

Size

Group Description

Noncardiac chest pain

Stable angina pectoris

Unstable angina pectoris

Myocardial infarction (heart attack)

Sample

Mean

35

35

35

35

Sample Standard

Deviation

10.89

11.25

11.37

11.75

0.69

0.74

0.91

1.07

With mi denoting the true mean MPV for group i (i ϭ 1, 2, 3, 4), let’s consider

the null hypothesis H0: m1 ϭ m2 ϭ m3 ϭ m4. Figure 15.2 shows a comparative boxplot for the four samples (based on data consistent with summary values given in the

paper). The mean MPV for the heart attack sample is larger than for the other three

samples and the boxplot for the heart attack sample appears to be shifted a bit higher

Noncardiac

Stable angina

Unstable angina

Myocardial infarction

FIGURE 15.2

Boxplots for Example 15.1.

9

10

11

MPV

12

13

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15.1

Single-Factor ANOVA and the F Test

707

than the boxplots for the other three samples. However, because the four boxplots

show substantial overlap, it is not obvious whether H0 is true or false. In situations

such as this, we need a formal test procedure.

As with the inferential methods of previous chapters, the validity of the ANOVA

test for H0: m1 5 m2 5 c5 mk requires some assumptions.

Assumptions for ANOVA

1. Each of the k population or treatment response distributions is normal.

2. s1 5 s2 5 % 5 sk (The k normal distributions have identical standard

deviations.)

3. The observations in the sample from any particular one of the k populations or treatments are independent of one another.

4. When comparing population means, the k random samples are selected

independently of one another. When comparing treatment means, treatments are assigned at random to subjects or objects (or subjects are

assigned at random to treatments).

In practice, the test based on these assumptions works well as long as the assumptions are not too badly violated. If the sample sizes are reasonably large, normal probability plots or boxplots of the data in each sample are helpful in checking the assumption of normality. Often, however, sample sizes are so small that a separate

normal probability plot or boxplot for each sample is of little value in checking normality. In this case, a single combined plot can be constructed by first subtracting x1

from each observation in the first sample, x2 from each value in the second sample,

and so on and then constructing a normal probability or boxplot of all N deviations

from their respective means. The plot should be reasonably straight. Figure 15.3

shows such a normal probability plot for the data of Example 15.1.

13

Residual

12

11

10

9

FIGURE 15.3

A normal probability plot using the

combined data of Example 15.1.

−3

−2

−1

0

Normal score

1

2

3

There is a formal procedure for testing the equality of population standard deviations. Unfortunately, it is quite sensitive to even a small departure from the normality

assumption, so we do not recommend its use. Instead, we suggest that the ANOVA

F test (to be described later in this section) can safely be used if the largest of the

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708

Chapter 15 Analysis of Variance

sample standard deviations is at most twice the smallest one. The largest standard

deviation in Example 15.1 is s4 ϭ 1.07, which is only about 1.5 times the smallest

standard deviation (s1 ϭ 0.69). The book Beyond ANOVA: The Basics of Applied Statistics by Rupert (see the references in the back of the book) is a good source for alternative methods of analysis if there appears to be a violation of assumptions.

The analysis of variance test procedure is based on the following measures of

variation in the data.

DEFINITION

A measure of differences among the sample means is the treatment sum of

squares, denoted by SSTr and given by

SSTr 5 n1 1x1 2 x 2 2 1 n2 1x2 2 x 2 2 1 % 1 nk 1xk 2 x 2 2

A measure of variation within the k samples, called error sum of squares and

denoted by SSE, is

SSE 5 1n1 2 12 s21 1 1n2 2 12 s22 1 % 1 1nk 2 12 s2k

Each sum of squares has an associated df:

treatment df 5 k 2 1

error df 5 N 2 k

A mean square is a sum of squares divided by its df. In particular,

SSTr

k21

SSE

mean square for error 5 MSE 5

N2k

mean square for treatments 5 MSTr 5

The number of error degrees of freedom comes from adding the number of degrees of freedom associated with each of the sample variances:

1n1 2 12 1 1n2 2 12 1 % 1 1nk 2 12 5 n1 1 n2 1 % 1 nk 2 1 2 1 2 % 2 1

5N2k

EXAMPLE 15.2

Heart Attack Calculations

Let’s return to the mean platelet volume (MPV) data of Example 15.1. The grand

mean x was computed to be 11.315. Notice that because the sample sizes are all

equal, the grand mean is just the average of the four sample means (this will not usually be the case when the sample sizes are unequal). With x1 ϭ 10.89, x2 ϭ 11.25,

x3 ϭ 11.37, x4 ϭ 11.75, and n1 ϭ n2 ϭ n3 ϭ n4 ϭ 35,

SSTr 5 n1 1x1 2 x 2 2 1 n2 1x2 2 x 2 2 1 % 1 nk 1xk 2 x 2 2

5 35 110.89 2 11.3152 2 1 35 111.25 2 11.3152 2 1 35 111.37 2 11.3152 2

1 35 111.75 2 11.3152 2

5 6.322 1 0.148 1 0.106 1 6.623

5 13.199

Because s1 ϭ 0.69, s2 ϭ 0.74, s3 ϭ 0.91, and s4 ϭ 1.07

SSE 5 1n1 2 12 s 21 1 1n2 2 12 s 22 1 % 1 1nk 2 12 s 2k

5 135 2 12 10.692 2 1 135 2 12 10.742 2 1 135 2 12 10.912 2 1 135 2 12 11.072 2

5 101.888

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15.1

Single-Factor ANOVA and the F Test

709

The numbers of degrees of freedom are

treatment df ϭ k Ϫ 1 ϭ 3

error df ϭ N Ϫ k ϭ 35 ϩ 35 ϩ 35 ϩ 35 Ϫ 4 ϭ 136

from which

MSTr 5

MSE 5

13.199

SSTr

5

5 4.400

k21

3

SSE

101.888

5

5 0.749

N2k

136

Both MSTr and MSE are quantities whose values can be calculated once sample

data are available; that is, they are statistics. Each of these statistics varies in value

from data set to data set. Both statistics MSTr and MSE have sampling distributions,

and these sampling distributions have mean values. The following box describes the

key relationship between MSTr and MSE and the mean values of these two

statistics.

When H0 is true 1m1 5 m2 5 % 5 mk2 ,

mMSTr 5 mMSE

However, when H0 is false,

mMSTr . mMSE

and the greater the differences among the m’s, the larger mMSTr will be relative to mMSE.

According to this result, when H0 is true, we expect the two mean squares to be

close to one another, whereas we expect MSTr to substantially exceed MSE when

some m’s differ greatly from others. Thus, a calculated MSTr that is much larger than

MSE casts doubt on H0. In Example 15.2, MSTr ϭ 4.400 and MSE ϭ 0.749, so

MSTr is about 6 times as large as MSE. Can this difference be attributed solely to

sampling variability, or is the ratio MSTr/MSE large enough to suggest that H0 is

false? Before we can describe a formal test procedure, it is necessary to revisit F distributions, first introduced in multiple regression analysis (Chapter 14).

Many ANOVA test procedures are based on a family of probability distributions

called F distributions. An F distribution always arises in connection with a ratio. A

particular F distribution is obtained by specifying both numerator degrees of freedom

(df1) and denominator degrees of freedom (df2). Figure 15.4 shows an F curve for a

particular choice of df1 and df2. All F tests in this book are upper-tailed, so P-values

are areas under the F curve to the right of the calculated values of F.

F curve for particular df1, df2

Shaded area = P-value for upper-tailed F test

FIGURE 15.4

An F curve and P-value for an uppertailed test.

Calculated F

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710

Chapter 15

Analysis of Variance

Tabulation of these upper-tail areas is cumbersome, because there are two degrees

of freedom rather than just one (as in the case of t distributions). For selected (df1,

df2) pairs, the F table (Appendix Table 6) gives only the four numbers that capture

tail areas .10, .05, .01, and .001, respectively. Here are the four numbers for df1 ϭ 4,

df2 ϭ 10 along with the statements that can be made about the P-value:

Tail area

Value

a.

b.

c.

d.

e.

.10

2.61

a

b

.05

3.48

c

.01

5.99

d

.001

11.28

e

F Ͻ 2.61 → tail area ϭ P-value Ͼ .10

2.61 Ͻ F Ͻ 3.48 → .05 Ͻ P-value Ͻ .10

3.48 Ͻ F Ͻ 5.99 → .01 Ͻ P-value Ͻ .05

5.99 Ͻ F Ͻ 11.28 → .001 Ͻ P-value Ͻ .01

F Ͼ 11.28 → P-value Ͻ .001

If F ϭ 7.12, then .001 Ͻ P-value Ͻ .01. If a test with a Ͻ .05 is used, H0 should be

rejected, because P-value Յ a. The most frequently used statistical computer packages can provide exact P-values for F tests.

The Single-Factor ANOVA F Test

Null hypothesis:

Test statistic:

H 0: m 1 5 m 2 5 % 5 m k

F5

MSTr

MSE

When H0 is true and the ANOVA assumptions are reasonable, F has an F distribution with df1 5 k 2 1 and df2 5 N 2 k.

Values of F more inconsistent with H0 than what was observed in the data are

values even farther out in the upper tail, so the P-value is the area captured in

the upper tail of the corresponding F curve. Appendix Table 6, a statistical

software package, or a graphing calculator can be used to determine P-values for

F tests.

\ EXAMPLE 15.3

Heart Attacks Revisited

The two mean squares for the MPV data given in Example 15.1 were calculated in

Example 15.2 as

MSTr ϭ 4.400

MSE ϭ 0.749

The value of the F statistic is then

F5

4.400

MSTr

5

5 5.87

MSE

0.749

with df1 ϭ k Ϫ 1 ϭ 3 and df2 ϭ N Ϫ k ϭ 140 Ϫ 4 ϭ 136. Using df1 ϭ 3 and

df2 ϭ 120 (the closest value to 136 that appears in the table), Appendix Table 6

shows that 5.78 captures tail area .001. Since 5.87 Ͼ 5.78, it follows that P-value ϭ

captured tail area Ͻ .001. The P-value is smaller than any reasonable a, so there is

compelling evidence for rejecting H0: m1 5 m2 5 % 5 m4. We can conclude that the

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15.1

Single-Factor ANOVA and the F Test

711

mean MPV is not the same for all four patient populations. Techniques for determining which means differ are introduced in Section 15.2.

EXAMPLE 15.4

Hormones and Body Fat

The article “Growth Hormone and Sex Steroid Administration in Healthy Aged

Women and Men” (Journal of the American Medical Association [2002]: 2282–

2292) described an experiment to investigate the effect of four treatments on various

body characteristics. In this double-blind experiment, each of 57 female subjects age

65 or older was assigned at random to one of the following four treatments: (1) placebo “growth hormone” and placebo “steroid” (denoted by P ϩ P); (2) placebo

“growth hormone” and the steroid estradiol (denoted by P ϩ S); (3) growth hormone

and placebo “steroid” (denoted by G ϩ P); and (4) growth hormone and the steroid

estradiol (denoted by G ϩ S).

The following table lists data on change in body fat mass over the 26-week period

following the treatments that are consistent with summary quantities given in the

article.

CHANGE IN BODY FAT MASS (KG)

Treatment

n

x

s

s2

P؉P

P؉S

G؉P

G؉S

0.1

0.6

2.2

0.7

Ϫ2.0

0.7

0.0

Ϫ2.6

Ϫ1.4

1.5

2.8

0.3

Ϫ1.0

Ϫ1.0

Ϫ0.1

0.2

0.0

Ϫ0.4

Ϫ0.9

Ϫ1.1

1.2

0.1

0.7

Ϫ2.0

Ϫ0.9

Ϫ3.0

1.0

1.2

Ϫ1.6

Ϫ0.4

0.4

Ϫ2.0

Ϫ3.4

Ϫ2.8

Ϫ2.2

Ϫ1.8

Ϫ3.3

Ϫ2.1

Ϫ3.6

Ϫ0.4

Ϫ3.1

0.014

0.064

1.545

2.387

0.014

Ϫ0.286

1.218

1.484

0.013

Ϫ2.023

1.264

1.598

Ϫ3.1

Ϫ3.2

Ϫ2.0

Ϫ2.0

Ϫ3.3

Ϫ0.5

Ϫ4.5

Ϫ0.7

Ϫ1.8

Ϫ2.3

Ϫ1.3

Ϫ1.0

Ϫ5.6

Ϫ2.9

Ϫ1.6

Ϫ0.2

0.016

Ϫ2.250

1.468

2.155

265.4

5 21.15.

57

Let’s carry out an F test to see whether actual mean change in body fat mass differs for the four treatments.

Also, N ϭ 57, grand total ϭ Ϫ65.4, and x 5

Step-by-Step technology

instructions available online

1. Let m1, m2, m3, and m4 denote the true mean change in body fat for treatments

P ϩ P, P ϩ S, G ϩ P, and G ϩ S, respectively.

Data set available online

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712

Chapter 15

Analysis of Variance

2. H0: m1 ϭ m2 ϭ m3 ϭ m4

3. Ha: At least two among m1, m2, m3, and m4 are different.

4. Significance level: a ϭ .01

5. Test statistic: F 5

MSTr

MSE

6. Assumptions: Figure 15.5 shows boxplots of the data from each of the four

samples. The boxplots are roughly symmetric, and there are no outliers. The largest standard deviation (s1 ϭ 1.545) is not more than twice as big as the smallest

(s2 ϭ 1.264). The subjects were randomly assigned to treatments. The assumptions of ANOVA are reasonable.

P+P

P+S

G+P

G+S

FIGURE 15.5

−6

−5

−4

Boxplots for the data of Example 15.4.

−3 −2 −1

0

1

Change in body fat mass

2

3

7. Computation:

SSTr 5 n1 1x1 2 x 2 2 1 n2 1x2 2 x 2 2 1 % 1 nk 1xk 2 x 2 2

5 14 10.064 2 121.152 2 2 1 14 120.286 2 121.152 2 2

1 13 122.023 2 121.152 2 2 1 16 122.250 2 121.152 2 2

5 60.37

treatment df ϭ k Ϫ 1 ϭ 3

SSE 5 1n1 2 12 s 21 1 1n2 2 12 s 22 1 % 1 1nk 2 12 s 2k

5 13 12.3872 1 13 11.4842 1 12 11.5982 1 15 12.1552

5 101.81

error df 5 N 2 k 5 57 2 4 5 53

Thus,

F5

MSTr

SSTr/treatment df

60.37 /3

20.12

5

5

5

5 10.48

MSE

SSE/error df

101.81 /53

1.92

8. P-value: Appendix Table 6 shows that for df1 ϭ 3 and df2 ϭ 60 (the closest

tabled df to df 5 53), the value 6.17 captures upper-tail area .001. Because F ϭ

10.48 Ͼ 6.17, it follows P-value Ͻ .001.

9. Conclusion: Since P-value Յ a, we reject H0. The mean change in body fat mass

is not the same for all four treatments.

Summarizing an ANOVA

ANOVA calculations are often summarized in a tabular format called an ANOVA

table. To understand such a table, we must define one more sum of squares.

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15.1

Single-Factor ANOVA and the F Test

713

Total sum of squares, denoted by SSTo, is given by

SSTo 5 a 1x 2 x 2 2

all N obs.

with associated df 5 N 2 1.

The relationship between the three sums of squares SSTo, SSTr, and SSE is

SSTo 5 SSTr 1 SSE

which is called the fundamental identity for single-factor ANOVA.

The quantity SSTo, the sum of squared deviations about the grand mean, is a

measure of total variability in the data set consisting of all k samples. The quantity

SSE results from measuring variability separately within each sample and then combining as indicated in the formula for SSE. Such within-sample variability is present

regardless of whether or not H0 is true. The magnitude of SSTr, on the other hand,

has much to do with whether the null hypothesis is true or false. The more the m’s

differ from one another, the larger SSTr will tend to be. Thus, SSTr represents variation that can (at least to some extent) be explained by any differences between means.

An informal paraphrase of the fundamental identity for single-factor ANOVA is

total variation ϭ explained variation ϩ unexplained variation

Once any two of the sums of squares have been calculated, the remaining one is

easily obtained from the fundamental identity. Often SSTo and SSTr are calculated

first (using computational formulas given in the online appendix to this chapter), and

then SSE is obtained by subtraction: SSE ϭ SSTo Ϫ SSTr. All the degrees of freedom, sums of squares, and mean squares are entered in an ANOVA table, as displayed

in Table 15.2. The P-value usually appears to the right of F when the analysis is done

by a statistical software package.

T A B L E 15 .2 General Format for a Single-Factor ANOVA Table

Source of

Variation

df

Sum of Squares

Treatments

kϪ1

SSTr

Error

NϪk

SSE

Total

NϪ1

SSTo

F

Mean Square

SSTr

k21

SSE

MSE 5

N2k

MSTr 5

F5

MSTr

MSE

An ANOVA table from Minitab for the change in body fat mass data of Example

15.4 is shown in Table 15.3. The reported P-value is .000, consistent with our previous conclusion that P-value Ͻ .001.

T A B L E 15 .3 An ANOVA Table from Minitab for the Data of Example 15.4

One-way ANOVA

Source

Factor

Error

Total

DF

3

53

56

SS

60.37

101.81

162.18

MS

20.12

1.92

F

10.48

P

0.000

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714

Chapter 15 Analysis of Variance

EX E RC I S E S 1 5 . 1 - 1 5 . 1 3

15.1 Give as much information as you can about the

P-value for an upper-tailed F test in each of the following

situations.

a. df1 ϭ 4, df2 ϭ 15, F ϭ 5.37

b. df1 ϭ 4, df2 ϭ 15, F ϭ 1.90

c. df1 ϭ 4, df2 ϭ 15, F ϭ 4.89

d. df1 ϭ 3, df2 ϭ 20, F ϭ 14.48

e. df1 ϭ 3, df2 ϭ 20, F ϭ 2.69

f. df1 ϭ 4, df2 ϭ 50, F ϭ 3.24

15.2 Give as much information as you can about the

P-value of the single-factor ANOVA F test in each of

the following situations.

a. k ϭ 5, n1 ϭ n2 ϭ n3 ϭ n4 ϭ n5 ϭ 4, F ϭ 5.37

b. k ϭ 5, n1 ϭ n2 ϭ n3 ϭ 5, n4 ϭ n5 ϭ 4, F ϭ 2.83

c. k ϭ 3, n1 ϭ 4, n2 ϭ 5, n3 ϭ 6, F ϭ 5.02

d. k ϭ 3, n1 ϭ n2 ϭ 4, n3 ϭ 6, F ϭ 15.90

e. k ϭ 4, n1 ϭ n2 ϭ 15, n3 ϭ 12, n4 ϭ 10, F ϭ 1.75

15.3 Employees of a certain state university system can

choose from among four different health plans. Each

plan differs somewhat from the others in terms of hospitalization coverage. Four samples of recently hospitalized

individuals were selected, each sample consisting of

people covered by a different health plan. The length of

the hospital stay (number of days) was determined for

each individual selected.

a. What hypotheses would you test to decide whether

mean length of stay was related to health plan?

(Note: Carefully define the population characteristics

of interest.)

b. If each sample consisted of eight individuals and the

value of the ANOVA F statistic was F ϭ 4.37, what

conclusion would be appropriate for a test with

a ϭ .01?

c. Answer the question posed in Part (b) if the F value

given there resulted from sample sizes n1 ϭ 9,

n2 ϭ 8, n3 ϭ 7, and n4 ϭ 8.

15.4 The accompanying summary statistics for a measure of social marginality for samples of youths, young

ceived Causes of Loneliness in Adulthood” (Journal of

Social Behavior and Personality [2000]: 67–84). The

social marginality score measured actual and perceived

social rejection, with higher scores indicating greater social rejection. For purposes of this exercise, assume that it

is reasonable to regard the four samples as representative

of the U.S. population in the corresponding age groups

Data set available online

and that the distributions of social marginality scores for

these four groups are approximately normal with the

same standard deviation. Is there evidence that the mean

social marginality score is not the same for all four age

groups? Test the relevant hypotheses using a 5 .01.

Age Group

Youths

Young

Seniors

Sample Size

x

s

106

2.00

1.56

255

3.40

1.68

314

3.07

1.66

36

2.84

1.89

15.5

The authors of the paper “Age and Violent

Content Labels Make Video Games Forbidden Fruits

for Youth” (Pediatrics [2009]: 870–876) carried out an

experiment to determine if restrictive labels on video

games actually increased the attractiveness of the game for

young game players. Participants read a description of a

new video game and were asked how much they wanted

to play the game. The description also included an age

rating. Some participants read the description with an age

restrictive label of 71, indicating that the game was not

appropriate for children under the age of 7. Others read

the same description, but with an age restrictive label of

121, 161, or 181. The data below for 12- to 13-year-old

boys are fictitious, but are consistent with summary statistics given in the paper. (The sample sizes in the actual

experiment were larger.) For purposes of this exercise, you

can assume that the boys were assigned at random to one

of the four age label treatments (71, 121, 161, and

181). Data shown are the boys’ ratings of how much they

wanted to play the game on a scale of 1 to 10. Do the data

provide convincing evidence that the mean rating associated with the game description by 12- to 13-year-old boys

is not the same for all four restrictive rating labels? Test the

appropriate hypotheses using a significance level of .05.

71 label

121 label

161 label

181 label

6

6

6

5

4

8

6

1

2

4

8

7

8

5

7

9

5

8

4

7

7

9

8

6

7

4

8

9

6

7

10

9

6

8

7

6

8

9

10

8

Video Solution available

Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).

Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

15.1

The paper referenced in the previous exercise

also gave data for 12- to 13-year-old girls. Data consistent with summary values in the paper are shown below.

Do the data provide convincing evidence that the mean

rating associated with the game description for 12- to

13-year-old girls is not the same for all four age restrictive rating labels? Test the appropriate hypotheses using

a 5 .05.

15.6

71 label

121 label

161 label

181 label

4

7

6

5

3

6

4

5

10

5

4

5

4

6

3

5

3

8

5

9

6

4

8

6

10

8

6

6

8

5

8

6

6

5

7

4

10

6

8

7

The paper “Women’s and Men’s Eating Behavior Following Exposure to Ideal-Body Images and

Text” (Communication Research [2006]: 507–529)

Single-Factor ANOVA and the F Test

715

Treatment 1 Treatment 2 Treatment 3 Treatment 4

1

5

8

11

5

1

0

6

4

10

7

0

12

8

6

2

7

8

8

5

14

9

0

6

3

12

5

6

10

8

6

2

10

0

3

4

4

5

5

7

8

4

0

6

3

2

0

0

3

4

2

4

1

1

15.7

describes an experiment in which 74 men were assigned

at random to one of four treatments:

1. Viewed slides of fit, muscular men

2. Viewed slides of fit, muscular men accompanied by

diet and fitness-related text

3. Viewed slides of fit, muscular men accompanied by

text not related to diet and fitness

4. Did not view any slides

The participants then went to a room to complete a

questionnaire. In this room, bowls of pretzels were set

out on the tables. A research assistant noted how many

pretzels were consumed by each participant while completing the questionnaire. Data consistent with summary

quantities given in the paper are given in the accompanying table. Do these data provide convincing evidence

that the mean number of pretzels consumed is not the

same for all four treatments? Test the relevant hypotheses

using a significance level of .05.

15.8 Can use of an online plagiarism-detection system

reduce plagiarism in student research papers? The paper

“Plagiarism and Technology: A Tool for Coping with

Plagiarism” (Journal of Education for Business [2005]:

149–152) describes a study in which randomly selected research papers submitted by students during five semesters

were analyzed for plagiarism. For each paper, the percentage

of plagiarized words in the paper was determined by an

online analysis. In each of the five semesters, students were

told during the first two class meetings that they would have

to submit an electronic version of their research papers and

that the papers would be reviewed for plagiarism. Suppose

that the number of papers sampled in each of the five semesters and the means and standard deviations for percentage of plagiarized words are as given in the accompanying

table. For purposes of this exercise, assume that the conditions necessary for the ANOVA F test are reasonable. Do

these data provide evidence to support the claim that mean

percentage of plagiarized words is not the same for all five

semesters? Test the appropriate hypotheses using a 5 .05.

Semester

n

Mean

Standard deviation

1

2

3

4

5

39

42

32

32

34

6.31

3.31

1.79

1.83

1.50

3.75

3.06

3.25

3.13

2.37

Treatment 1 Treatment 2 Treatment 3 Treatment 4

8

7

4

13

2

6

8

0

4

9

1

5

2

0

3

5

2

5

7

5

(continued)

Data set available online

Video Solution available

Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).

Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

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